The t-interval

The same grammar, two substitutions

When $\sigma$ is unknown, replace it by $s$ and replace $z_{\alpha /2}$ by $t_{\alpha /2;n-1}$:

$$\bar{x}\pm t_{\alpha /2;n-1}\frac{s}{\sqrt{n}}$$

Everything else — estimate ± critical × SE, the interpretation, the width drivers — carries over from L9.1 and L9.2 unchanged. The only new work is the table lookup: row $n-1$, column $\alpha /2$.

Worked example — Cedar & Stone's billable hours

Cedar & Stone Accounting wants the mean billable hours per client file. A random sample of $n=20$ files has $\bar{x}=4.85$ hours and $s=1.10$ hours; no history of $\sigma$ exists. A 90 % interval: $\alpha =0.10$, so the column is 0.05 and the row is df $=19$:

$$t_{0.05;19}=1.729,\mathrm{SE}=\frac{1.10}{\sqrt{20}}=0.246,\text{margin}=1.729\times 0.246=0.425$$

$$4.85\pm 0.425⟹(4.42,5.28)\text{ hours}$$

A second engagement: the firm's audit fee per small-business client, $n=10$ files, $\bar{x} = \$212.4$,$s = \$18.6$, at 95 %. Column 0.025, row 9: $t_{0.025;\,9} = 2.262$.

$$\mathrm{SE}=\frac{18.6}{\sqrt{10}}=5.882,\text{margin}=2.262\times 5.882=13.30,(199.10,225.70)$$

With only ten files the multiplier is 2.262 instead of 1.96 — the interval is 15 % wider than a z-interval would be, which is honest: with $\sigma$ estimated from nine degrees of freedom, the analyst knows less.

The follow-up the paper asks

How would the method change if 1.10 were the known population SD? Then $z$ replaces $t$: the 90 % multiplier becomes $1.645$, the margin $1.645\times 0.246=0.405$, and the interval narrows to $(4.45,5.25)$. Knowing $\sigma$ removes the uncertainty that fattened the tails. The data and the SE are the same in both; the critical value is the whole difference — and the decision between them is made by whether $\sigma$ is given, never by the size of $n$.

When t is not safe

The t procedure assumes the population is roughly normal, because for a small sample the CLT has not yet made $\bar{X}$ normal on its own. The course's robustness guideline: the t-interval is reliable for $n\ge 15$ unless the sample shows outliers or strong skew, and for $n\ge 40$ regardless of shape. Below 15, or with a visibly skewed plot, check a histogram or box plot (Module 3) and say what you see. Cedar & Stone's $n=20$ passes if the hours look reasonably symmetric; the ten audit fees are on the edge and the analyst should say so.

The common failure is the reverse one: using $z=1.645$ with $s$ and $n=20$ because "the sample is big enough". Twenty is not $\infty$; the interval comes out too narrow and the marker deducts. Use $t$ whenever $\sigma$ is estimated and $n\le 30$; L9.5 covers what happens beyond 30.